On the Circumcenters of Cevasix Configurations
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1 Forum Geometricorum Volume 3 (2003) FORUM GEOM ISSN On the ircumcenters of evasix onfigurations lexei Myakishev and Peter Y. Woo bstract. We strengthen Floor van Lamoen s theorem that the 6 circumcenters of the cevasix configuration of the centroid of a triangle are concyclic by giving a proof which at the same time shows that the converse is also true with a minor qualification, i.e., the circumcenters of the cevasix configuration of a point P are concyclic if and only if P is the centroid or the orthocenter of the triangle. 1. Introduction Let P be a point in the plane of triangle, with traces,, on the sidelines,, respectively. We assume that P does not lie on any of the sidelines. Triangle is then divided by its cevians,, into six triangles, giving rise to what lark Kimberling [2, pp ] called the cevasix configuration of P. See Figure 1. Floor van Lamoen has discovered that when P is the centroid of triangle, the 6 circumcenters of the cevasix configuration are concyclic. See Figure 2. This was posed as a problem in the merican Mathematical Monthly [3]. Solutions can be found in [3, 4]. In this note we study the converse. + ( +) ( ) G ( ) ( +) G F ( +) ( ) + + Figure 1 Figure 2 Theorem 1. The circumcenters of the cevasix configuration of P are concyclic if and only if P is the centroid or the orthocenter of triangle. Publication Date: March 3, ommunicating Editor: Paul Yiu.
2 58. Myakishev and P. Y. Woo 2. Preliminary results We adopt the following notations. Triangle P P P P P P Notation ( + ) ( ) ( + ) ( ) ( + ) ( ) ircumcenter It is easy to see that two of these triangle may possibly share a common circumcenter only when they share a common vertex of triangle. Lemma 2. The circumcenters of triangles P and P coincide if and only if P lies on the reflection of the circumcircle in the line. Proof. Triangles P and P have the same circumcenter if and only if the four points,, P, are concyclic. In terms of directed angles, P = P = = =. See, for example, [1, 16 20]. It follows that the reflection of in the line lies on the circumcircle of triangle P, and P lies on the reflection of the circumcircle in. The converse is clear. Thus, if + = and + =, then necessarily P is the orthocenter H, and also + =. In this case, there are only three distinct circumcenters. They clearly lie on the nine-point circle of triangle. We shall therefore assume P H, so that there are at least five distinct points in the set { ±, ±, ± }. The next proposition appears in [2, p.259]. Proposition 3. The 6 circumcenters of the cevasix configuration of P lie on a conic. Proof. We need only consider the case when these 6 circumcenters are all distinct. The circumcenters + and lie on the perpendicular bisector of the segment P ; similarly, and + lie on the perpendicular bisector of P. These two perpendicular bisectors are clearly parallel. This means that + and + are parallel. Similarly, + // + and + // +. The hexagon has three pairs of parallel opposite sides. y the converse of Pascal s theorem, there is a conic passing through the six vertices of the hexagon. Proposition 4. The vertices of a hexagon with parallel opposite sides + // +, + // +, + // + lie on a circle if and only if the main diagonals +, + and + have equal lengths. Proof. If the vertices are concyclic, then + + is an isosceles trapezoid, and + = +. Similarly, + + is also an isosceles trapezoid, and + = +. onversely, consider the triangle XY Z bounded by the three diagonals +, + and +. If these diagonals are equal in length, then the trapezoids + +, + + and + + are isosceles. From these we immediately conclude that the common perpendicular bisector of + and +
3 On the circumcenters of cevasix configurations 59 is the bisector of angle XY Z. Similarly, the common perpendicular bisector of + and + is the bisector of angle X, and that of + and + the bisector of angle Z. These three perpendicular bisectors clearly intersect at a point, the incenter of triangle XY Z, which is equidistant from the six vertices of the hexagon. Proposition 5. The vector sum + + = 0 if and only if P is the centroid. Proof. Suppose with reference to triangle, the point P has absolute barycentric coordinates u + v + w, where u + v + w =1. Then, = 1 v + w (v + w), = 1 w + u (w + u), = 1 (u + v). u + v From these, + + =( + + ) ( + + ) u 2 vw = (w + u)(u + v) + v 2 wu (u + v)(v + w) + w 2 uv (v + w)(w + u). This is zero if and only if u 2 vw = v 2 wu = w 2 uv =0, and u = v = w = 1 3 since they are all real, and u + v + w =1. We denote by π a, π b, π c the orthogonal projections on the lines,, respectively. Proposition 6. π b ( + )= 1 2, π c ( + )= 1 2, π c ( + )= 1 2, π a ( + )= 1 2, (1) π a ( + )= 1 2, π b ( + )= 1 2. Proof. The orthogonal projections of + and on the cevian are respectively the midpoints of the segments P and P. Therefore, π b ( + )= + P 2 The others follow similarly. 3. Proof of Theorem 1 P + 2 = 2 = 1 2. Sufficiency part. Let P be the centroid G of triangle. y Proposition 4, it is enough to prove that the diagonals +, + and + have equal lengths. y Proposition 5, we can construct a triangle whose sides as vectors, and are equal to the medians,, respectively.
4 60. Myakishev and P. Y. Woo onsider the vector Q equal to +. y Proposition 6, the orthogonal projections of + on the two sides and are the midpoints of the sides. This means that Q is the circumcenter of triangle, and the length of + is equal to the circumradius of triangle. The same is true for the lengths of + and +. The case P = H is trivial. Necessity part. Suppose the 6 circumcenters ±, ±, ± lie on a circle. y Proposition 3, the diagonals +, +, and + have equal lengths. We show that + + =0, so that P is the centroid of triangle by Proposition 5. In terms of scalar products, we rewrite equation (1) as + = 1 2, + = 1 2, + = 1 2, + = 1 2, (2) + = 1 2, + = 1 2. From these, ( ) =0, and is orthogonal to Since +, and + have equal lengths, and + + are orthogonal. We may therefore write + + = k for a scalar k. From (2) above, k =( + + ) = =. From this, k =1and we have = + +. The same reasoning shows that = + +, = + +. ombining the three equations, we have + + = 0. It follows from Proposition 5 that P must be the centroid of triangle. 4. n alternative proof of Theorem 1 We present another proof of Theorem 1 by considering an auxiliary hexagon. Let L a and L a be the lines perpendicular to at and respectively; similarly, L b, L b, and L c and L c. onsider the points X + = L c L b, X = L b L c, Y + = L a L c, Y = L c L a, Z + = L b L a, Z = L a L b.
5 On the circumcenters of cevasix configurations 61 Note that the circumcenters ±, ±, ± are respectively the midpoints of PX ±, PY ±, PZ ±. Hence, the six circumcenters are concyclic if and only if X ±, Y ±, Z ± are concyclic. In Figure 3, let P = P = α. Since angles P Y and PY are both right angles, the four points P,,, Y are concyclic and Z + Y X + = Y X + = P = α. Similarly, P = P = Y X + Z, and we denote the common measure by β. Y + Z α β α P β α α Y X β β Z + X + Figure 3 Lemma 7. If the four points X +, Y, Z +, Z are concyclic, then P lies on the median through. Proof. Let x = P and y = P. If the four points X +, Y, Z +, Z are concyclic, then Z + Z X + = α and Y Z + Z = β. Now, = Z +Z sin α Z + Z sin β = sin α sin β = It follows that P = P, and, as a ratio of signed lengths, P P = y x. (3).
6 62. Myakishev and P. Y. Woo Now applying Menelaus theorem to triangle P with transversal, and triangle G with transversal,wehave P P = 1 = P P. From this, P = P,or 1 x = 1 y. (4) omparing (3) and (4), we have 1 x 1 y = y x, (x y)(x + y 1) = 0. Either x = y or x + y =1. It is easy to eliminate the possibility x + y =1.IfPhas homogeneous barycentric coordinates (u : v : w) with reference to triangle, then x = v+w u+v+w and y = w+u u+v+w. Thus, x + y =1requires w =0and P lies on the sideline, contrary to the assumption. It follows that x = y, and from (3), is the midpoint of, and P lies on the median through. The necessity part of Theorem 1 is now an immediate corollary of Lemma oncluding remark We conclude with a remark on triangles for which two of the circumcenters of the cevasix configuration of the centroid coincide. learly, + = if and only if,, G, are concyclic. Equivalently, the image of G under the homothety h(, 2) lies on the circumcircle of triangle. This point has homogeneous barycentric coordinates ( 1 :2:2). Since the circumcircle has equation a 2 yz + b 2 zx + c 2 xy =0, we have 2a 2 = b 2 + c 2. There are many interesting properties of such triangles. We simply mention that it is similar to its own triangle of medians. Specifically, m a = 2 a, m b = 2 c, m c = 2 b. Editor s endnote. John H. onway [5] has located the center of the Van Lamoen circle (of the circumcenters of the cevasix configuration of the centroid) as F = mn + cot2 ω (G K), 12 where mn is the medial Ninecenter, 1 G the centroid, K the symmedian point, and ω the rocard angle of triangle. In particular, the parallel through F to the symmedian line GK hits the Euler line in mn. See Figure 4. The point F has homogeneous barycentric coordinates (10a 4 13a 2 (b 2 + c 2 )+(4b 4 10b 2 c 2 +4c 4 ): : ). This appears as X 1153 of [6]. 1 This is the point which divides OH in the ratio 1:3.
7 On the circumcenters of cevasix configurations 63 + K N mn G F O H + + Figure 4 References [1] R.. Johnson, dvanced Euclidean Geometry, 1925, Dover reprint. [2]. Kimberling, Triangle centers and central triangles, ongressus Numerantium, 129 (1998) [3] F. M. van Lamoen, Problem 10830, mer. Math. Monthly, 2000 (107) 863; solution by the Monthly editors, 2002 (109) [4] K. Y. Li, oncyclic problems, Mathematical Excalibur, 6 (2001) Number 1, 1 2; available at [5] J. H. onway, Hyacinthos message 5555, May 24, [6]. Kimberling, Encyclopedia of Triangle enters, ugust 22, 2002 edition, available at February 17, 2003 edition available at lexei Myakishev: Smolnaia 61-2, 138, Moscow, Russia, address: alex geom@mtu-net.ru Peter Y. Woo: Department of Mathematics, iola University, iola venue, La Mirada, alifornia 90639, US address: woobiola@yahoo.com
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